Showing posts with label IIM Calcutta. Show all posts
Showing posts with label IIM Calcutta. Show all posts

Thursday, January 12, 2012

CAT 2011 Results

CAT 2011 results are out and I have managed a decent looking 99.78%ile with a 99.86 in QA and 95.43 in VA.

I have managed  calls from IIM C and L as of now, and hopefully I will manage a few more.

I was expecting a score of 75-80 in quant and around 45-50 in verbal.

An overall of 120-130

Thursday, September 29, 2011

Problem of the Day 29 Sept 2011

Let a, b, c be three distinct odd natural numbers. Which of the following can be the sum of the squares of a, b and c?

(a) 3333                            (b) 5555                                (c) 9999                        (d) 7777




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Friday, September 23, 2011

Problem of the Day 23 Sept 2011


Darryl has a six-sided die with faces 1; 2; 3; 4; 5; 6. He knows the die is weighted so that one face
comes up with probability 1/2 and the other fi ve faces have equal probability of coming up. He
unfortunately does not know which side is weighted, but he knows each face is equally likely
to be the weighted one.

He rolls the die 5 times and gets a 1; 2; 3; 4 and 5 in some unspecifi ed order. Compute the probability that his next roll is a 6.



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Thursday, September 22, 2011

How to use the options to maximize scores in QA/DI/LR in IIM CAT

The good thing about CAT is, that it is not a mathematics test. So, it helps that you know the process and the steps but you should be smart and try to minimize the work. This is why there are options to help you. A smart use of options will save a lot of time and help you increase your score.

Some examples

Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?


(a) 70         (b) 71                (c) 72                         (d) 73


Now it is pretty much obvious the number is of the form 55xy
To find xy we can just subtract 55 from the options which gives xy= 15, 16, 17, 18

If you are lucky you will start with 73 and you will know it is right, else even if you start with 15, you will soon reach 18 and get your answer.


The direct method will be a bit long.

 Let n be the total number of different 5-digit numbers with all distinct digits, formed using 2, 3, 4, 5 and 6 and divisible by 4. What is the value of n?

1] 44 2] 32 3] 36 4] 38 5] 40

Permutations & Combinations is probably the most ‘hated’ topic. However, if you understand the basics and use logic, then it is the most fun-filled topic of all. Let us get to this question. As mentioned, we need to find out the 5 digit numbers divisible by 4 formed by the digits given in the question. To be divisible by 4, the last 2 digits should be divisible by 4. So to arrive at the answer, the first step is to find out combinations of the 2 digits from 2,3,4,5 & 6 that are divisible by 4 – eg: 24. Then, for each such combination, the last 2 digits are fixed. The remaining 3 digits can be arranged in 3! Ways = 6 ways. So the answer would be - 6 multiplied by the total no. of combinations of 2 digits divisible by 4. The answer necessarily should be a multiple of 6 and therefore the answer is 36 – option [3]. We just got lucky here, by the way, since there is just one option that is divisible by 6!



Problem of the Day 22 Sept 2011


Susan plays a game in which she rolls two fair standard six-sided dice with sides labeled one through six. She wins if the number on one of the dice is three times the number on the other die. If Susan plays this game three times, compute the probability that she wins at least once.